COMP SCI 5400 SP2017 Exam 1 Key

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1 COMP SCI 5400 SP2017 Exam 1 Key This is a closed-book, closed-notes exam. The only items you are permitted to use are writing implements. Mark each sheet of paper you use with your name and the string cs5400 sp2017 exam1. If you are caught cheating, you will receive a zero grade for this exam. The max number of points per question is indicated in square brackets after each question. The sum of the max points is 75, but the max exam score will be capped at 72 (i.e., there are 3 bonus points, but you can t score more than 100%). You have 75 minutes to complete this exam. Good luck! 1. For each of the following PES categories, from the comma separated lists of properties, indicate the correct property for this semester s puzzle (you do not have to justify your answer): (a) Fully Observable, Partially Observable [1] (b) Deterministic, Stochastic, Strategic [1] (c) Episodic, Sequential [1] (d) Static, Dynamic, Semi-dynamic [1] (e) Discrete, Continuous [1] (f) Single gent, Multi-agent [1] (g) Known, Unknown [1] 2. Using the corresponding letter labels, give a formal description of this semester s puzzle by defining the: (a) Initial State [3] - n m n grid with the upper left corner denoted as the coordinate pair (0,0) and the bottom right corner as (m 1,n 1), one cell containing the back of the air boat and one adjacent cell containing the front of the air boat, one cell representing the escape location, a set of movable objects in the form of giant mutant alligators (each alligator occupies a 1 by 3 grid cell rectangle with an orientation) and giant mutant turtles (each turtle occupies a 1 by 2 grid cell rectangle with an orientation), a set of stationary objects in the form of Cypress trees (each tree is represented by a single grid cell), and a radiation source grid cell with its associated radiation magnitude and decay factor such that all cells have positive radiation costs; the air boat, movable objects, and stationary objects must all occupy unique (i.e., non-overlapping) cells. (b) ctions Function [3] - CTIONS(s) returns the set of valid actions in state s which consists of all single moves of either the air boat or one of the mutant creatures as long as they remain entirely on the grid at all times and not move into or through any occupied cell. The air boat can move one cell forward or turn 90 degrees left or right from its back axis. The mutant creatures can only move a single cell forward or backward. 1

2 (c) Transition Model [3] - RESULT(s,a) returns the successor state of s resulting from performing action a which specifies that exactly one movable object has been moved as follows depending on the object type: air boat in s: (back cell, front cell) = ((x b, y b ), (x f, y f )), in successor state: forward orientation: oriented up : ((x b, y b 1), (x f, y f 1)) oriented down : ((x b, y b + 1), (x f, y f + 1)) oriented left : ((x b 1, y b ), (x f 1, y f )) oriented right : ((x b + 1, y b ), (x f + 1, y f )) rotate left orientation: oriented up : ((x b, y b ), (x f 1, y f + 1)) oriented down : ((x b, y b ), (x f + 1, y f 1)) oriented left : ((x b, y b ), (x f + 1, y f + 1)) oriented right : ((x b, y b ), (x f 1, y f 1)) rotate right orientation: oriented up : ((x b, y b ), (x f + 1, y f + 1)) oriented down : ((x b, y b ), (x f 1, y f 1)) oriented left : ((x b, y b ), (x f + 1, y f 1)) oriented right : ((x b, y b ), (x f 1, y f + 1)) turtle in s: (back cell, front cell) = ((x b, y b ), (x f, y f )), in successor state: forward identical to air boat backward orientation: oriented up : ((x b, y b + 1), (x f, y f + 1)) oriented down : ((x b, y b 1), (x f, y f 1)) oriented left : ((x b + 1, y b ), (x f + 1, y f )) oriented right : ((x b 1, y b ), (x f 1, y f )) alligator identical to turtle, except for three cells rather than two cells (d) Goal Test [1] - Returns true when the front of the air boat occupies the goal cell representing the escape location; otherwise false. (e) Path Cost Function [1] - The sum of the step costs of the actions to get to the current state from the initial state, where step cost is defined as the sum of the radiation incurred by the air boat over the two cells it occupies at the end of each turn, where cell radiation is defined by the radiation magnitude of the source cell minus the Manhattan distance between the cell and the source cell multiplied with the decay factor. In a turn where the goal is reached, no radiation is incurred. 3. Given two admissible heuristics h 1 and h 2. (a) What does it mean for h 1 to dominate h 2? [2] h 1 dominates h 2 iff ( n, h 1 (n) h 2 (n) n, h 1 (n) > h 2 (n)) (b) If h 1 dominates h 2, what is the implication for *TS using h 1 versus *TS using h 2? [2] If h 1 dominates h 2, then *TS using h 1 will never expand more nodes than *TS using h 2, except possibly for some nodes with f(n) = C (c) re there any circumstances under which it would be beneficial to include in the max composite heuristic, two heuristics of which one is dominated by the other? Justify your answer! [2] No, because the max function will always select the dominant heuristic value which is guaranteed to never be worse than the dominated heuristic value and the computational time required to compute the dominated heuristic is therefore always wasted. 2

3 The next questions are about the following state space graph, with being the start state and I being the only goal state. The edge labels indicate step-cost, the vertex labels contain the node identifier in the form of a letter. Heuristic h(node) is defined as the minimum number of steps from node to the goal state; for example, h() = 4. The order in which successors are generated is counterclockwise, ending at exactly 9 o clock. Example: E generates first H, then F, then B, and finally D. When sorting by path-cost, nodes with equal path-cost are ordered such that the earlier a node is generated, the higher its priority. Nodes already on the frontier have higher priority than newly added nodes with equal path-cost. You may use the following abbreviations without defining them: GF = Goal Found, NGF = No Goal Found, DLR = Depth Limit Reached. 4. Give the execution trace for Uniform Cost Graph Search (UCGS). [8] 0 0 D2 B2 D2 B2 G5 E5 D B2 C3 G5 E5 D B C3 G5 E5 F5 D B C G5 E5 F5 H10 D B C G E5 F5 H7 D B C G E F5 H7 I11 D B C G E F H7 I8 D B C G E F H I8 goal found; solution = DEHI; path-cost(dehi) = 8 5. Is UCGS optimal for this problem? Justify your answer! [3] Yes, because the branching factor is finite and any path originating in the start state which includes actions with non-positive step costs must go through the goal state before encountering said actions, and UCGS will terminate once it encounters a goal state, and hence the step costs of all reachable actions are positive, and UCGS is known to be optimal under those conditions. 6. If actions with arbitrary step costs are added to this state space graph, would UCGS still be guaranteed complete for the resulting state space? Justify your answer. [2] No, because it would, for instance, get stuck following an infinite sequence of zero-cost actions. 3

4 7. Give the execution trace for Iterative Deepening Depth First Graph Search (ID-DFGS). [12] depth-limit=0 depth-limit=1 D B D B D B depth-limit=2 D B D G E B D G E B D G E B D G E B C D G E B C depth-limit=3 D B D G E B D G H E B D G H E B D G H E F B D G H E F B D G H E F B C D G H E F B C depth-limit=4 D B D G E B D G H E B D G H I E B D G H I goal found; solution = DGHI; path-cost(dghi) = Is ID-DFGS optimal for this problem? Justify your answer! [1] No, because it found a solution with a higher cost (11) than the one found by UCGS (8). 4

5 9. Give the execution trace for Graph Search ( GS) employing heuristic h. [10] 4 4 D5 B5 D5 B5 G7 E7 D B5 C5 G7 E7 D B C5 F6 G7 E7 D B C F6 G7 E7 I11 D B C F G7 E7 I11 H11 D B C F G E7 H8 I11 D B C F G E H8 I8 D B C F G E H I8 goal found; solution = DEHI; path-cost(dehi) = If the step cost to go from I to J and from J to I were 1 rather than 0, would for this problem h be consistent? Justify your answer! [7] Yes, because (I) all the step costs would be one or larger (c(n, a, n ) 1) and (II) the minimal number of steps to the goal from a node and from its successor would differ by exactly one ( h(n) h(n ) = 1); applying constraint (I) to the definition of consistency gives n, n : h(n) 1 + h(n ) c(n, a, n ) + h(n ) and applying constraint (II) to the left-hand side of this equation gives n, n : h(n) h(n ) 1 = h(n) h(n ) which is always true. 11. If the step cost to go from I to J and from J to I were 1 rather than 0, would for this problem h be admissible? Justify your answer! [3] Yes, because the previous answer showed it to be consistent. 12. Is GS employing heuristic h optimal for this problem? Justify your answer! [1] Yes, because it found the same solution as UCGS which we previously stated to be optimal for this problem. 13. Define heuristic h (node) as the maximum number of steps from node to the goal state without visiting the same node twice; for example, h () = 8. Now define the composite heuristic h c (node) as max{h(node), h (node)}. If the step cost to go from I to J and from J to I were 1 rather than 0, would for this problem h c be consistent? Justify your answer! [4] No, because n, h (n) h(n) so h c (n) = h (n) and h (n) is not consistent because it s not admissible because h (C) = 8 > 7 = C (C). 5